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Diagonalizabilty versus spread for uncertainty (discrete)

  1. Dec 31, 2014 #1


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    I have seen two characterizations of the problem in measuring a discrete variable of a state ψ exactly with each of two non-commuting Hermitian operators A and B:
    (1) that the product of the standard deviations ( = √(<ψ|A2|ψ>-<ψ|A|ψ>2), & ditto for B) ≥ 1
    (2) that one cannot simultaneously diagonalize the matrix representations of A and B
    (i.e., if A = UCU and B = VDV, for unitary U and V and diagonal C and D, with denoting the adjoint, then U≠V.
    Where is the link between these two?
  2. jcsd
  3. Jan 1, 2015 #2


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    As explained here: http://en.wikipedia.org/wiki/Uncert...2.80.93Schr.C3.B6dinger_uncertainty_relations, the link is that

    [itex]\sigma_A \sigma_B \geq \frac{1}{2} | \langle [A,B] \rangle|[/itex]
  4. Jan 1, 2015 #3


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    Thanks, stevendaryl. Light is starting to seep in, as I read in the link you cited "... in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding orthonormal bases in Hilbert space are Fourier transforms of one another ..... A nonzero function and its Fourier transform cannot both be sharply localized." So I apparently have to delve into the theory of Fourier transforms next.
    Forgive me for appearing a bit dense (a poor imitation of the Tortoise in Lewis Carroll's "What the Tortoise said to Achilles") , but I do not see that the link (between the HUP and the non-simultaneous-diagonalizability complementarity) is the more general case (than my use of "1" in units of ħ/2) of the HUP which you stated , since in looking at the derivation it appears to me that all that is required to show that |<[ A,B]>| is non-zero is the non-commutativity, and not the non-simultaneous-diagonalizability.
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